# Speed & Distance - Solved Examples

Q 1 - What is meters/sec for 54 km/hr?

A - 15 m/sec

B - 20 m/sec

C - 25 m/sec

D - 30 m/sec

Explanation

```54 km/hr = (54*5/18) m/sec = 15 m/sec.
```

Q 2 - What is km/hr for 16 m/sec?

A - 53.6km/hr

B - 55.6km/hr

C - 57.6km/hr

D - 59.6km/hr

Explanation

```16 m/sec = (16*18/5)km/hr = 288/5km/hr =57.6 km/hr.
```

Q 3 - Anita can cover a sure separation in 1 hr 24min by covering two-third of the separation at 4km/hr and the rest at5 km/hr. Find the aggregate separation.

A - 3 kms

B - 4 kms

C - 5 kms

D - 6 kms

Explanation

```Let the aggregate separation be x km.
Then, 2/3x/4+1/3x/5=7/5
=> x/6+x/15=7/5
=> 5x+2x=42
=> 7x=42
=> x=6.
∴ Total separation = 6 km
```

Q 4 - A man strolls from his home to the railroad station. On the off chance that he strolls at 5 km/hr, he misses a train by 7 minutes. However, on the off chance that he strolls at 6km/hr, he achieves the station 5 minute before the flight of the train. Discover the separation secured by him to achieve the station.

A - 5 kms

B - 6 kms

C - 7 kms

D - 8 kms

Explanation

```Let the required separation be x km. At that point,
X/5 - x/6 = 12/60 (distinction between two time interims is 12 min.)
=> x/5 - x/6 = 1/5
=> 6x-5x=6
=> x= 6
Required separation = 6 kms
```

Q 5 - Strolling at 7/8 of its typical velocity, a train is 10 minutes past the point of no return. Locate its standard time to cover the trip?

A - 60 min

B - 70 min

C - 80 min

D - 90 min

Explanation

```New speed = 7/8 of its standard velocity
New time taken = 8/7 of the standard time.
(8/7 of the standard time)- (common time) = 10 min.
=> 1/7 of the standard time = 10 min
=> usual time = 70 min.
```

Q 6 - Hitesh covers a sure separation via auto driving at 70 km/hr and returns back to the beginning stage riding on a bike at 55km/hr. locate his normal velocity for the entire trip?

A - 62.6 km/hr

B - 61.6 km/hr

C - 60.6 km/hr

D - 59.6 km/hr

Explanation

```Normal velocity = 2xy/(x+y) km/hr = (2*70*55)/ (70+55) km/hr
= (2*70*55)/125 km/hr = 308/5 km/hr = 61.6 km/hr
```

Q 7 - The separation between two stations A and B is 450 km. A train begins at 4 pm from A and moves towards B at a normal velocity of 60 km/hr. Another train begins from B at 3.20 p.m and moves towards A at a normal velocity of 80 km/hr. How a long way from A will the two train s meet and what time?

A - 6:50 pm

B - 5:50 pm

C - 4:50 pm

D - 3:50 pm

Explanation

```Assume two trains meet at x km from A
(time taken by B to cover (450-x) km-(time taken by A to cover x km) = 40/60
=> (450-x)/80 - x/60 = 40/60 ?3 (450-x) - 4x = 160
=> 7x=1190 ∴ x = 170
In this way the two trains meet at a separation of 170 km from A.
Time taken by A to cover 170 km = 170/60 = 2hrs 50 min.
Along these lines, the two trains meet at 6:50 pm
```

Q 8 - A man cycles from A to B, a separation of 21 km in 1 hr 40 min. The street from A is level for 13 km and afterward it is tough to B. The man's normal rate on level is 15 km/hr. Locate his normal tough pace?

A - 10 km/hr

B - 11 km/hr

C - 12 km/hr

D - 13 km/hr

Explanation

```Let the normal tough rate be x km/hr. at that point,
13/15 + 8/x = 5/3
=>8/x= (5/3-13/15) = 12/15 = 4/5
=>x = (8*5)/4 = 10
∴ Normal tough rate = 10 km/hr
```

Q 9 - A hoodlum is spotted by a policeman from a separation of 100 meters. At the point when the policeman begins the pursuit, the criminal likewise begin s running. In the event that the pace of the criminal be 8km/hr and that of the policeman 10 km/hr, how far the hoodlum will have keep running before he is overwhelmed?

A - 200 m

B - 300 m

C - 400 m

D - 500 m

Explanation

```Relative pace of the policeman = (10-8) km/hr = 2 km/hr
Time taken by policeman to cover 100 m = (100/1000*1/2) hr = 1/20 hr
In 1/20 hr, the cheat covers a separation of (8*1/20) km= (2/5 km) = (2/5*1000) m = 400 m
```

Q 10 - I walk a sure separation and ride back setting aside an aggregate time of 37 minutes. I could walk both routes in 55 minutes. To what extent would it take me to ride both ways?

A - 18 min

B - 19 min

C - 20 min

D - 21 min

Explanation

```Let the given separation be x km. At that point,
(Time taken to walk X km)+ (time taken to ride x km) = 37 min.
=> (time taken to walk 2X km) + (time taken to ride 2x km) = 74 min.
=> 55 min + (time taken to ride 2 x km) = 74
=> time taken to ride 2x km = 19 min.
```

Q 11 - A and B are two stations 390 km separated. A train begins from An at 10 am and ventures towards B at 65 kmph. Another train begins from B at 11 am and towards A at 35 kmph. AT what time they meet?

A - 2.15 pm

B - 3.15 pm

C - 4.15 pm

D - 5.15 pm

Explanation

```Assume they meet x hours after 10 am then,
(separation moved by first in x hrs.) + (separation moved by second in (x-1) hrs) = 390
∴ 65x+ 35(x-1) = 390
=> 100x =425
=> x= 17/4
In this way, they meet 4 hrs 15 min. after 10 am at 2.15 pm
```

Q 12 - A products train leaves a station at a sure time and at a settled rate. Following 6 hours, an express prepare leaves the same station and moves in the same bearing at a uniform velocity of 90 kmph. This train makes up for lost time the merchandise train in 4 hours. Discover the velocity of the merchandise train.

A - 33 kmph

B - 34 kmph

C - 35 kmph

D - 36 kmph

```Let the velocity of the merchandise train be x kmph.